Answer :

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

⇒ Dividend = Divisor × Quotient + Remainder

When x + 2 is factor of polynomial

Then

x + 2 = 0

x = – 2

if x + 2 is factor of polynomial f(x) = ax^{3} + bx^{2} + x – 6

then f( – 2) is 0

f( – 2) = a( – 2)^{3} + b( – 2)^{2} + ( – 2) – 6 = 0

– 8a + 4b – 2 – 6 = 0

4b – 8a = 8

4b = 8 + 8a ……… eq 1

When x – 2 divides polynomial gives remainder 4

Then

Dividend = Divisor × Quotient + Remainder

ax^{3} + bx^{2} + x – 6 = (x – 2) × Quotient + 4

ax^{3} + bx^{2} + x – 6 – 4 = (x – 2) × Quotient

ax^{3} + bx^{2} + x – 10 = (x – 2) × Quotient

(x – 2) = 0

x = 2

if (x – 2) is factor of polynomial ax^{3} + bx^{2} + x – 10

then f(2) = 0

f(2) = a(2)^{3} + b(2)^{2} + 2 – 10

8a + 4b – 8 = 0

8a + 4b = 8 ………eq 2

Putting value of 4b from eq 1 into eq 2

8a + (8 + 8a) = 8

16a + 8 = 8

16a = 8 – 8 = 0

a = 0

Putting value of ‘a’ in eq 1

4b = 8 + 8a

4b = 8 + 8 × 0

4b = 8

b = = 2

Conclusion.

∴ The value of a and b comes to be 0 and 2 respectively

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By applying RemaiWest Bengal Mathematics

By applying RemaiWest Bengal Mathematics

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By applying RemaiWest Bengal Mathematics

Applying RemaindeWest Bengal Mathematics

Applying RemaindeWest Bengal Mathematics

By applying RemaiWest Bengal Mathematics

By applying RemaiWest Bengal Mathematics

By applying RemaiWest Bengal Mathematics

Applying RemaindeWest Bengal Mathematics